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受欢迎的三角函数 >

5cos^2(θ)+3sqrt(2)sin(θ)-11/2 =0

解答

5 cos^{2} (θ) + 32 sin (θ) - \frac{11}{2} = 0

解答

θ = 0.14189 \dots + 2 πn, θ = π - 0.14189 \dots + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

+1

度数

θ = 8.13010 \dots^{\circ} + 36 0^{\circ} n, θ = 171.86989 \dots^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 36 0^{\circ} n, θ = 13 5^{\circ} + 36 0^{\circ} n

求解步骤

5 cos^{2} (θ) + 32 sin (θ) - \frac{11}{2} = 0

化简

5 cos^{2} (θ) + 32 sin (θ) - \frac{11}{2} : \frac{10 cos ^{2} ( θ ) + 6 2 sin ( θ ) - 11}{2}

5 cos^{2} (θ) + 32 sin (θ) - \frac{11}{2}

将项转换为分式:

5 cos^{2} (θ) = \frac{5 cos ^{2} ( θ ) 2}{2}, 32 sin (θ) = \frac{3 \cdot 2 sin ( θ ) 2}{2}

= \frac{5 cos ^{2} ( θ ) \cdot 2}{2} + \frac{3 2 sin ( θ ) \cdot 2}{2} - \frac{11}{2}

因为分母相等，所以合并分式:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{5 cos ^{2} ( θ ) \cdot 2 + 3 2 sin ( θ ) \cdot 2 - 11}{2}

5 cos^{2} (θ) \cdot 2 + 32 sin (θ) \cdot 2 - 11 = 10 cos^{2} (θ) + 62 sin (θ) - 11

5 cos^{2} (θ) \cdot 2 + 32 sin (θ) \cdot 2 - 11

数字相乘:

5 \cdot 2 = 10

= 10 cos^{2} (θ) + 3 \cdot 22 sin (θ) - 11

数字相乘:

3 \cdot 2 = 6

= 10 cos^{2} (θ) + 62 sin (θ) - 11

= \frac{10 cos ^{2} ( θ ) + 6 2 sin ( θ ) - 11}{2}

\frac{10 cos ^{2} ( θ ) + 6 2 sin ( θ ) - 11}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

10 cos^{2} (θ) + 62 sin (θ) - 11 = 0

使用三角恒等式改写

- 11 + 10 cos^{2} (θ) + 6 sin (θ) 2

使用毕达哥拉斯恒等式:

cos^{2} (x) + sin^{2} (x) = 1

cos^{2} (x) = 1 - sin^{2} (x)

= - 11 + 10 (1 - sin^{2} (θ)) + 6 sin (θ) 2

化简

- 11 + 10 (1 - sin^{2} (θ)) + 6 sin (θ) 2 : 62 sin (θ) - 10 sin^{2} (θ) - 1

- 11 + 10 (1 - sin^{2} (θ)) + 6 sin (θ) 2

= - 11 + 10 (1 - sin^{2} (θ)) + 62 sin (θ)

乘开

10 (1 - sin^{2} (θ)) : 10 - 10 sin^{2} (θ)

10 (1 - sin^{2} (θ))

使用分配律:

a (b - c) = ab - a c

a = 10, b = 1, c = sin^{2} (θ)

= 10 \cdot 1 - 10 sin^{2} (θ)

数字相乘:

10 \cdot 1 = 10

= 10 - 10 sin^{2} (θ)

= - 11 + 10 - 10 sin^{2} (θ) + 6 sin (θ) 2

数字相加/相减:

- 11 + 10 = - 1

= 62 sin (θ) - 10 sin^{2} (θ) - 1

= 62 sin (θ) - 10 sin^{2} (θ) - 1

- 1 - 10 sin^{2} (θ) + 6 sin (θ) 2 = 0

用替代法求解

- 1 - 10 sin^{2} (θ) + 6 sin (θ) 2 = 0

令

: sin (θ) = u

- 1 - 10 u^{2} + 6 u 2 = 0

- 1 - 10 u^{2} + 6 u 2 = 0 : u = \frac{2}{10}, u = \frac{2}{2}

- 1 - 10 u^{2} + 6 u 2 = 0

改写成标准形式

a x^{2} + b x + c = 0

- 10 u^{2} + 62 u - 1 = 0

使用求根公式求解

- 10 u^{2} + 62 u - 1 = 0

二次方程求根公式:

若

a = - 10, b = 62, c = - 1

u_{1, 2} = \frac{- 6 2 \pm ( 6 2 ) ^{2} - 4 ( - 10 ) ( - 1 )}{2 ( - 10 )}

u_{1, 2} = \frac{- 6 2 \pm ( 6 2 ) ^{2} - 4 ( - 10 ) ( - 1 )}{2 ( - 10 )}

(62)^{2} - 4 (- 10) (- 1) = 42

(62)^{2} - 4 (- 10) (- 1)

使用法则

- (- a) = a

= (62)^{2} - 4 \cdot 10 \cdot 1

(62)^{2} = 6^{2} \cdot 2

(62)^{2}

使用指数法则:

(a \cdot b)^{n} = a^{n} b^{n}

= 6^{2} (2)^{2}

(2)^{2} : 2

使用根式运算法则:

a = a^{\frac{1}{2}}

= (2^{\frac{1}{2}})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= 2^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 \cdot 2}{2}

约分:

2

= 1

= 2

= 6^{2} \cdot 2

4 \cdot 10 \cdot 1 = 40

4 \cdot 10 \cdot 1

数字相乘:

4 \cdot 10 \cdot 1 = 40

= 40

= 6^{2} \cdot 2 - 40

6^{2} \cdot 2 = 72

6^{2} \cdot 2

6^{2} = 36

= 36 \cdot 2

数字相乘:

36 \cdot 2 = 72

= 72

= 72 - 40

数字相减:

72 - 40 = 32

= 32

32

质因数分解:

2^{5}

32

32

除以

232 = 16 \cdot 2

= 2 \cdot 16

16

除以

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 8

8

除以

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 4

4

除以

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

2

是质数，因此无法进一步因数分解

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

= 2^{5}

= 2^{5}

使用指数法则:

a^{b + c} = a^{b} \cdot a^{c}

= 2^{4} \cdot 2

使用根式运算法则:

n ab = n a n b

= 2 2^{4}

使用根式运算法则:

n a^{m} = a^{\frac{m}{n}}

2^{4} = 2^{\frac{4}{2}} = 2^{2}

= 2^{2} 2

整理后得

= 42

u_{1, 2} = \frac{- 6 2 \pm 4 2}{2 ( - 10 )}

将解分隔开

u_{1} = \frac{- 6 2 + 4 2}{2 ( - 10 )}, u_{2} = \frac{- 6 2 - 4 2}{2 ( - 10 )}

u = \frac{- 6 2 + 4 2}{2 ( - 10 )} : \frac{2}{10}

\frac{- 6 2 + 4 2}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 6 2 + 4 2}{- 2 \cdot 10}

同类项相加:

- 62 + 42 = - 22

= \frac{- 2 2}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 2 2}{- 20}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2 2}{20}

约分:

2

= \frac{2}{10}

u = \frac{- 6 2 - 4 2}{2 ( - 10 )} : \frac{2}{2}

\frac{- 6 2 - 4 2}{2 ( - 10 )}

去除括号:

(- a) = - a

= \frac{- 6 2 - 4 2}{- 2 \cdot 10}

同类项相加:

- 62 - 42 = - 102

= \frac{- 10 2}{- 2 \cdot 10}

数字相乘:

2 \cdot 10 = 20

= \frac{- 10 2}{- 20}

使用分式法则:

\frac{- a}{- b} = \frac{a}{b}

= \frac{10 2}{20}

约分:

10

= \frac{2}{2}

二次方程组的解是:

u = \frac{2}{10}, u = \frac{2}{2}

u = sin (θ)

代回

sin (θ) = \frac{2}{10}, sin (θ) = \frac{2}{2}

sin (θ) = \frac{2}{10}, sin (θ) = \frac{2}{2}

sin (θ) = \frac{2}{10} : θ = arcsin (\frac{2}{10}) + 2 πn, θ = π - arcsin (\frac{2}{10}) + 2 πn

sin (θ) = \frac{2}{10}

使用反三角函数性质

sin (θ) = \frac{2}{10}

sin (θ) = \frac{2}{10}

的通解

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{2}{10}) + 2 πn, θ = π - arcsin (\frac{2}{10}) + 2 πn

θ = arcsin (\frac{2}{10}) + 2 πn, θ = π - arcsin (\frac{2}{10}) + 2 πn

sin (θ) = \frac{2}{2} : θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

sin (θ) = \frac{2}{2}

sin (θ) = \frac{2}{2}

的通解

sin (x)

周期表（周期为

2 πn

"）：

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

合并所有解

θ = arcsin (\frac{2}{10}) + 2 πn, θ = π - arcsin (\frac{2}{10}) + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

以小数形式表示解

θ = 0.14189 \dots + 2 πn, θ = π - 0.14189 \dots + 2 πn, θ = \frac{π}{4} + 2 πn, θ = \frac{3 π}{4} + 2 πn

作图

流行的例子

cos^2(x)-5cos(x)+2=0

cos^{2} (x) - 5 cos (x) + 2 = 0

2sin^2(t)cos(t)=0

2 sin^{2} (t) cos (t) = 0

5sin(x)-12cos(x)=13

5 sin (x) - 12 cos (x) = 13

6 sin (θ) = 1

2cos((3x-pi)/6)+1=0,0<x<2pi

2 cos (\frac{3 x - π}{6}) + 1 = 0, 0 < x < 2 π